The r.m.s speed of oxygen molecules at 47°C$47^{\circ}\text{C}$ is equal to that of the hydrogen molecules kept at ________ °C$^{\circ}\text{C}$. (Mass of oxygen molecule / mass of hydrogen molecule = 32 / 2)
TH₂ in °C = 20 - 273 = -253°C$$T_{H_2} \text{ in }^{\circ}\text{C} = 20 - 273 = -253^{\circ}\text{C}$$
Pattern Recognition
Direct proportionality between Temperature and Molecular Mass for equal RMS speeds. T₁/M₁ = T₂/M₂$T_1/M_1 = T_2/M_2$. Always convert to Kelvin before evaluating.
Keywords:#r.m.s speed of oxygen molecules#JEE Main 2026 Evening Q30#Kinetic Theory of Gases JEE Main 2026#RMS Speed JEE Main 2026
More Kinetic Theory of Gases Previous-Year Questions
Q37jee_main_2026_22_january_morningRotating Gas Cylinder Pressure
A cylindrical tube AB of length l, closed at both ends contains an ideal gas of 1 mol having molecular weight M. The tube is rotated in a horizontal plane with constant angular velocity ω$\omega$ about an axis perpendicular to AB and passing through the edge at end A, as shown in the figure. If PA$P_{A}$ and PB$P_{B}$ are the pressures at A and B respectively, then (Consider the temperature is same at all points in the tube)
Cylindrical tube rotating about an axis passing through end A.
Sees: Rotating gas column with centrifugal pressure gradient.
Shortcut: Integrate hydrostatic equation with centripetal acceleration term.
Check: Matches option (1). ✓
Chapter Mix
Class 11 Physics: Kinetic Theory of Gases
Q28jee_main_2026_22_january_eveningMean Free Path and Collision Frequency
Consider two boxes containing ideal gases A and B such that their temperatures, pressures and number densities are same. The molecular size of A is half of that of B and mass of molecule A is four times that of B. If the collision frequency in gas B is 32 × 10¹⁸/s$32 \times 10^{18}/\mathrm{s}$ then collision frequency in gas A is ____/s.
A.32 × 10⁸$32 \times 10^{8}$
B.4 × 10⁸$4 \times 10^{8}$
C.2 × 10⁸$2 \times 10^{8}$
D.8 × 10⁸$8 \times 10^{8}$
Solution
Related Formula
Z = √(2)π d² N √((8RT)/(π M))$$Z = \sqrt{2}\pi d^{2} N \sqrt{\frac{8RT}{\pi M}}$$
where d$d$ is molecular diameter, N$N$ is number density, T$T$ is temperature, and M$M$ is molar mass.
Core Logic
Given that T$T$ and N$N$ are identical for both gases:
Z ∝ d²√(M)$$Z \propto \frac{d^2}{\sqrt{M}}$$
From the problem statement:
dA = (dB)/(2), MA = 4MB$$d_A = \frac{d_B}{2}, \quad M_A = 4M_B$$
An air bubble of volume 2.9 cm³$2.9 \, \mathrm{cm}^{3}$ rises from the bottom of a swimming pool of 5 m deep. At the bottom of the pool water temperature is 17 °C$17 \, ^{\circ}\mathrm{C}$ . The volume of the bubble when it reaches the surface, where the water temperature is 27 °C$27 \, ^{\circ}\mathrm{C}$ , is ____ cm³$\mathrm{cm}^{3}$ .
(g = 10 m/s²$g = 10 \, \mathrm{m/s}^{2}$ , density of water = 10³ kg/m³$10^{3} \, \mathrm{kg/m}^{3}$ , and 1 atm pressure is 10⁵ Pa$10^{5} \, \mathrm{Pa}$ )
For an air bubble rising in water, the number of moles of gas remains constant.
P₁$P_1$ = Pressure at bottom = Patm + ρ gh$P_{\text{atm}} + \rho gh$T₁$T_1$ = Temperature at bottom = 17°C = 290 K$17^{\circ}\mathrm{C} = 290 \, \mathrm{K}$V₁$V_1$ = 2.9 cm³$2.9 \, \mathrm{cm}^3$
P₂$P_2$ = Pressure at surface = Patm$P_{\text{atm}}$T₂$T_2$ = Temperature at surface = 27°C = 300 K$27^{\circ}\mathrm{C} = 300 \, \mathrm{K}$
Standard combined gas law application. Always remember to convert Celsius to Kelvin and carefully compute gauge pressure plus atmospheric pressure for the submerged state.
Chapter Mix
Class 11 Physics: Kinetic Theory of Gases
Class 11 Physics: Mechanical Properties of Fluids
Q47jee_main_2026_24_january_morningIdeal Gas Laws
A gas of certain mass filled in a closed cylinder at a pressure of 3.23 kPa has temperature 50°C$50^{\circ}\text{C}$. The gas is now heated to double its temperature. The modified pressure is ____ Pa.
Numerical Answer.Answer: 3730 to 3730
Solution
Related Formula
P ∝ T (Gay-Lussac's Law for constant V)$$P \propto T \quad (\text{Gay-Lussac's Law for constant V})$$
Core Logic
Since the gas is filled in a closed cylinder, the volume V$V$ remains constant.
Therefore, P ∝ T$P \propto T$ (where T must be in Kelvin).
Initial temperature Tᵢ = 50°C = 273 + 50 = 323 K$T_i = 50^{\circ}\text{C} = 273 + 50 = 323 \text{ K}$.
Final temperature Tf = 100°C = 273 + 100 = 373 K$T_f = 100^{\circ}\text{C} = 273 + 100 = 373 \text{ K}$.
Note: The phrasing "double its temperature" is slightly ambiguous (Celsius vs Kelvin). As per standard interpretations in such exams, "double its temperature" given in Celsius means 100°C$100^{\circ}\text{C}$ (which is 373 K). If it meant doubling absolute temperature, final would be 646 K. The official interpretation used here doubles the Celsius scale value.
Be highly alert to 'double temperature' traps when given in Celsius. The question specifically intends 50^° arrow 100^°$50^\circ \rightarrow 100^\circ$, not 323K arrow 646K$323\text{K} \rightarrow 646\text{K}$. Calculating with 373K nicely cancels with the 3.23kPa$3.23\text{kPa}$.
Chapter Mix
Class 11 Physics: Kinetic Theory
Class 11 Physics: Thermodynamics
More Kinetic Theory of Gases Questions — jee_main_2026_21_jan_evening
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
